First slide

Simple hormonic motion

Question

A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is $U (x) = k {[x]}^{3},$ where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

Difficult

A

$Proportional to \frac{1}{\sqrt{a}}$
B

Independent of a
C

Proportional to $\sqrt{a}$
D

Proportional to $a^{\frac{1}{2}}$

Solution

$U = k | x |^{3} \Rightarrow F = - \frac{dU}{dx} = - 3 k | x |^{2}$ ----------------(i)
Also, for SHM $x = a \sin ωt and \frac{d^{2} x}{{dt}^{2}} + ω^{2} x = 0$
$\Rightarrow acceleration = \frac{d^{2} x}{{dt}^{2}} = - ω^{2} x \Rightarrow F = ma$
$= m \frac{d^{2} x}{{dt}^{2}} = - {mω}^{2} x - - - - (ii)$
From equation (i) and (ii) we get $ω = \sqrt{\frac{3 kx}{m}}$
$\Rightarrow T = \frac{2 π}{ω} = 2 π \sqrt{\frac{m}{3 kx}} = 2 π \sqrt{\frac{m}{3 k (a \sin ωt)}} \Rightarrow T \propto \frac{1}{\sqrt{a}}$

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